For any bounded operator, \(C\) , define \(\mu _n(C)\)  by
\( \mu _n(C) = \min _{\psi _1 \dots \psi _{n-1}} \,\sup _{\begin{array}{c} \Vert \varphi \Vert =1 \\ \varphi \perp \psi _1, \dots , \psi _{n-1}\end{array}} \Vert C\varphi \Vert \) 
By the min-max principle (see [1]}),
\( \lim _{n\rightarrow \infty }\, \mu _n(C) =\sup (\sigma _\text{\rm {ess}}(\vert C\vert ))\) 
and \(\mu _n(C)\)  are the singular values if \(C\in {\mathcal {I}}_\infty \) . In particular,
\( C\in {\mathcal {I}}_\infty \Leftrightarrow \lim _{n\rightarrow \infty }\, \mu _n(C)=0\) 
Let \(\wedge ^\ell ({\mathcal {H}})\)  be the antisymmetric tensor product (see [2]},
[1]}, and [4]}). As usual (see [4]}),
\( \Vert \wedge ^m (C)\Vert  = \prod _{j=1}^m \mu _j(C)\) 
Since \(\mu _1\ge \mu _2 \ge \cdots \ge 0\) , we have
\( \lim _{n\rightarrow \infty }\, \mu _n(C) =\lim _{n\rightarrow \infty }\, (\mu _1(C) \dots \mu _n(C))^{1/n}\) 
(REF )–(REF ) imply
\( C\in {\mathcal {I}}_\infty \Leftrightarrow \lim _{n\rightarrow \infty }\, \Vert \wedge ^n (C)\Vert ^{1/n} =0\) 
As usual, there is a selfadjoint operator, \(d\wedge ^n(A)\)  on \(\wedge ^n({\mathcal {H}})\)  so
\( \wedge ^n (e^{-tA})=e^{-t\, d\wedge ^n(A)}\) 
so Segal's lemma implies that
\(\Vert \wedge ^n (e^{-(A+B)})\Vert  &\le \Vert \wedge ^n (e^{-A}) \wedge ^n (e^{-B})\Vert  \\&= \Vert \wedge ^n (e^{-A} e^{-B})\Vert  \) 
Thus,
\( \lim _{n\rightarrow \infty }\, \Vert \wedge ^n (e^{-(A+B)})\Vert ^{1/n} \le \lim _{n\rightarrow \infty }\,\Vert \wedge ^n (e^{-A} e^{-B})\Vert ^{1/n}\) 
By (REF ), we obtain (REF ).
